Classwork Series and Exercises {Mathematics- JSS2}: Binary System (Operations and Applications)

Number bases

Most people count in tens. For instance, the place value of the digits in number 4956 in this system

Thousands     hundreds    tens      units

       ↓                       ↓                   ↓             ↓

       4                    9                5             6

The place values are in powers of ten. It is called the base ten system.

4956 = 4 X 1000 + 9 X 100 + 5 X 10 + 6 X 1

          = 4 X10³ + 9 X 10² + 5 X 10¹ + 6 X 1⁰

Some people traditionally count in 5s, other in 20s. Using the method of the base ten system, a base five system is in powers of five. So, for example, in base five, 342 would be

342_five = 3 X 25 + 4 X 5 + 2 X 1

           = 3 X 5² + 4 X 5¹+ 2 X 5⁰

Notice that 342_five is short for 342 in base five.

Converting base ten numbers to other bases

To convert from base ten to another base, express the given number in powers of the new base.

Example

Convert 37_ten  a. to base eight,  b. to base five.

7. Since 37 < 64, there are no sixty-fours in 37. To find the number of eights in 37, divide by 8.

37 ÷ 8 = 4, remainder 5

        37 = 4 eights + 5 units

     37_ten = 45_eight

Check:  45_eight = 4 X 8 + 5 X 1

                         = 32 + 5

                         = 37

Since 37 > 25, there must be a twenty-five in 37.

37 ÷ 25 = 1, remainder 12

37 = 1 twenty-five + 12 units.

Consider the 12 units. Since 12 > 5, there must be some fives in 12.

12 ÷ 5 = 2, remainder 2

       12 = 2 fives + 2 units

   ∴ 37 = 1 twenty-five + 2 fives + 2 units

            = 1 x 5² + 2 X 5¹ + 2 X 1

   37_ten = 122_five

Check: 122_five = 1 X 25 + 2 X 5 + 2 X 1

                        = 25 + 10 + 2

                        = 37

The method in part b of Example 2 can be shortened as follows:

5 | 37

5 |   7 + 2 (i.e  7 X 5 + 2 X 1)

5 |  1 + 2  ( i.e. 1 X 5² + 2 X 5¹)

5 |  0 + 1  ( i.e. 0 X 5³ + 1 X 5 ²)

             ↑                           ↑

Continued division by 5 gives remainders. Reading the remainders upwards gives    37_ten = 122_five

(see the arrows above).

To change from base ten to another base:

1. Divide the base ten number by the new base number.
2. Continue dividing until 0 (zero) is reached, writing down the remainder each time.
3. Start at the last remainder and read upwards to get the answer.

Convert from other bases to base ten

Convert an Octal into a Decimal
This is a little bit of the same as before except our base is now 8 instead of 16.  Try it with the number 723₈ also written 0723 in computer programming notation.  Go ahead and convert it.  I’ll outline the steps below again.

723₈ = (7 X 8²) + (2 X 8¹) + (3 X 8⁰)
=    448 + 16 + 3
=  467₁₀

723 octal is 467 decimal.  See, that’s not so tough.  Now, the astute reader may have noticed another way this can be accomplished using the same information from the original formula.  Take the previous octal number as an example here. Since you know what the base and exponent will be for all digits’ place, you can go ahead and write that out: 82 = 64, 81 = 8, 80 = 1. Write your number beneath and multiply out, then add the products.

Operations with binary numbers

To add, subtract and multiply binary number, use the method that you use with base ten numbers. However, you must remember that you are working with powers of two, not powers of ten. The following identities are very useful:

Addition

0 + 0 = 0      1 + 0 = 1

0 + 1 = 2        1 + 1 = 10

Multiplication

0 X 0 = 0       1 X 0 = 0

0 X 1 =  0       1 X 1 = 1

The example below shows how to use these identities when operating with binary numbers. Follow the notes carefully.

QUESTIONS

 Lets see how much you’ve learnt, attach the following answers to the comment below

Calculate the following in base two:

1.      1 001

      + 1 011

—————

—————

2.       1 110

        –    101

—————–

—————–

3.           110

         x   101

—————–

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